91Ó°ÊÓ

The company's policy is to purchase a load when the factory rejects a pile is 2.1 or less.

The random variable x is defined as the number of flaws per piece.

Provided that random variable x has a Poisson distribution with a mean of 2.5.The probability that the contractor will not purchase a load of siding is calculated using the normal distribution.

The Poisson distribution has the same mean and variance.

Here$\stackrel{¯}{x}$is the average number of flaws having meanlocalid="1659736038381" $=2.5$and variance,localid="1659736032501" $=2.5$

Hence,

localid="1659736046989" ${\mathrm{μ}}_{\stackrel{¯}{x}}=2.5$,

localid="1659736043170" $\begin{array}{rcl}{\mathrm{σ}}_{\stackrel{¯}{x}}& =& \frac{\sqrt{2.5}}{\sqrt{35}}\\ & =& 0.2678\\ & & \end{array}$

The contractor will not purchase if the average number of flaws is 2.1 or more.The associated normal curve is drawn as follows:

Therefore, the required probability is,

$\begin{array}{rcl}P\left(\stackrel{¯}{x}â\copyright ¾2.1\right)& =& P\left(\frac{\stackrel{¯}{x}-{\mathrm{μ}}_{\stackrel{¯}{x}}}{{\mathrm{σ}}_{\stackrel{→}{x}}}â\copyright ¾\frac{2.1-2.5}{0.2673}\right)\\ & =& P\left(zâ\copyright ¾-1.4967\right)\\ & =& 0.5+A\\ & & \end{array}$

$$\begin{gathered} =0.5+0.4332 \\ =0.9332 \end{gathered}$$

Hence, the contractor's probability of not purchasing a load of siding is 0.9332.